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  • Gear ratio problem

    From none) (albert@21:1/5 to All on Wed Jan 4 10:01:30 2023
    I have a milling setup on my lathe. (Emco Maximat VP10)
    The motor spins at 1380 rpm.
    The spindle speeds are 350 640 780 and 1450.

    Derive a possible (probable) configuration of gears with
    their tooth count.
    Hints:
    do not expect gears with more than say 30 teeth

    [This is an approximate problem, if you had to calculate gears
    for thread cutting at the lathe, ratio's are fixed. ]

    Groetjes Albert
    --
    Don't praise the day before the evening. One swallow doesn't make
    spring. You must not say "hey" before you have crossed the bridge.
    Don't sell the hide of the bear until you shot it. Better one bird in
    the hand than ten in the air. - the Wise from Antrim -

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  • From Lorem Ipsum@21:1/5 to none albert on Wed Jan 4 11:08:29 2023
    On Wednesday, January 4, 2023 at 4:01:33 AM UTC-5, none albert wrote:
    I have a milling setup on my lathe. (Emco Maximat VP10)
    The motor spins at 1380 rpm.
    The spindle speeds are 350 640 780 and 1450.

    Derive a possible (probable) configuration of gears with
    their tooth count.
    Hints:
    do not expect gears with more than say 30 teeth

    [This is an approximate problem, if you had to calculate gears
    for thread cutting at the lathe, ratio's are fixed. ]

    Not sure what you meany by "an approximate problem". Are you saying you only need to be close?

    The motor speed is 1380 RPM. The first spindle speed is 350. Eliminating the common divisors yields these divisors, where each number is 30 or less.

    1380 - 6 and 23
    350 - 5 and 7
    So to keep the number of teeth below 30 in all cases, requires two sets of gears with ratios of 5:6 and 7:23.

    1380 - 3 and 23
    640 - 4 and 8
    So 4:3 and 8:23

    1380 - 23
    780 - 13
    So 13:23

    1380 - 6 and 23
    1450 - 5 and 29
    So 5:6 and 29:23

    With an algorithm for finding appropriate divisors, this could be automated. However, it's not a time intensive process to do it by hand. I think this took me 5 minutes, most of which was formatting the data from the web page to calculate the divisors.

    --

    Rick C.

    - Get 1,000 miles of free Supercharging
    - Tesla referral code - https://ts.la/richard11209

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  • From none) (albert@21:1/5 to [email protected] on Thu Jan 5 14:01:39 2023
    In article <[email protected]>,
    Lorem Ipsum <[email protected]> wrote:
    On Wednesday, January 4, 2023 at 4:01:33 AM UTC-5, none albert wrote:
    I have a milling setup on my lathe. (Emco Maximat VP10)
    The motor spins at 1380 rpm.
    The spindle speeds are 350 640 780 and 1450.

    Derive a possible (probable) configuration of gears with
    their tooth count.
    Hints:
    do not expect gears with more than say 30 teeth

    [This is an approximate problem, if you had to calculate gears
    for thread cutting at the lathe, ratio's are fixed. ]

    Not sure what you meany by "an approximate problem".
    Are you saying you only need to be close?

    Exactly.
    Suppose you have a gear ratio 19 to 20
    1380 20 19 */ .
    1452 OK
    Probably the manufacturer is likely to say 1450
    1380 42 40 */ .
    1449 OK
    Probably the manufacturer is likely to say 1450


    The motor speed is 1380 RPM. The first spindle speed is 350. Eliminating the common divisors yields these divisors, where each number is 30 or less.

    1380 - 6 and 23
    350 - 5 and 7
    So to keep the number of teeth below 30 in all cases, requires two sets of gears with ratios of 5:6 and 7:23.

    1380 - 3 and 23
    640 - 4 and 8
    So 4:3 and 8:23

    1380 - 23
    780 - 13
    So 13:23

    1380 - 6 and 23
    1450 - 5 and 29
    So 5:6 and 29:23

    With an algorithm for finding appropriate divisors, this could be automated. However, it's not a time intensive process to do it by hand. I think this took me 5 minutes, most of which was
    formatting the data from the web page to calculate the divisors.

    More information.
    the motor axis contains gears A1 A2
    An intermediate axis contains B1 and B2
    The spindle axis contains C1 and C2
    C2 has 42 teeth.
    That is the largest.
    Gears with less that 10 teeth are impractical.
    (Bicycle gears start with 10 teeth.)

    The engagement of the gears is
    1450 A2-B1 B2-C1
    780 A1-B1 B2-C1
    640 A2-B1 B1-C2 (! B1 twice)
    350 A1-B2 B1-C2

    There is two levers.
    You can see that one of the levers shift the intermediate axis
    from B1 to B2.
    The priority is that shifting gears is mechanical simplicity through
    shifting an assembly with two gears over an axis.
    It commands respect that it all works out.
    No synchro-mesh. (Nur im Stillstand schalten)

    With the additional information it is probably possible
    to find out to number of teeth.

    --

    Rick C.

    Groetjes Albert
    --
    Don't praise the day before the evening. One swallow doesn't make
    spring. You must not say "hey" before you have crossed the bridge.
    Don't sell the hide of the bear until you shot it. Better one bird in
    the hand than ten in the air. - the Wise from Antrim -

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    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From none) (albert@21:1/5 to albert on Fri Jan 6 11:04:27 2023
    In article <nnd$6e563037$12f16139@08b1045f14ba952c>,
    none) (albert <albert@cherry.> wrote:
    I have a milling setup on my lathe. (Emco Maximat VP10)
    The motor spins at 1380 rpm.
    The spindle speeds are 350 640 780 and 1450.

    Derive a possible (probable) configuration of gears with
    their tooth count.
    Hints:
    do not expect gears with more than say 30 teeth

    [This is an approximate problem, if you had to calculate gears
    for thread cutting at the lathe, ratio's are fixed. ]

    Finding an approximation for e.g. 350:1380 (milling setup)
    was supposed to be an example of continued fractions.

    1380 350 .CF
    3 1 16 2 |10
    This are the coefficients with 10 the greatest common divider.

    Going back is easy enough
    \\: 3 \\
    OK
    2DUP frac
    1/3 OK
    1 \\
    OK
    2DUP frac
    1/4 OK
    16 \\
    OK
    2DUP frac
    17/67 OK
    2 \\
    OK
    2DUP frac
    35/138 OK

    So feeding back the coefficients to \\: and \\ finds
    approximations with increasing precision.

    The words are not complicated.

    \ For A B print its continued fraction and the GCD.
    : .CF SWAP BEGIN OVER /MOD . DUP WHILE SWAP REPEAT DROP &| EMIT . ;

    \ Leave START value for convergents.
    : \\: 0 1 1 0 ;

    \ For CONV1 CONV2 incorporate N (cf term), return CONV2 CONV3
    : \\ >R 2SWAP 2OVER R@ * SWAP R> * SWAP D+ ;


    You know the coefficients of sqr(3) : 1 1 2 1 2 1 2 ..

    \\: 1 \\ 1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\
    1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\ 1 \\ 2 \\
    OK
    2DUP frac
    1542841/2672279 OK

    Now check it
    SQ SWAP SQ 2 + SWAP
    OK
    2DUP . .
    2380358351281 7141075053843 OK
    /MOD . .
    3 0

    The fraction was 2 units in 12 digits precise.

    [Leaving the printing of fraction and SQ to the imagination
    of the reader.]

    Groetjes Albert
    --
    Don't praise the day before the evening. One swallow doesn't make
    spring. You must not say "hey" before you have crossed the bridge.
    Don't sell the hide of the bear until you shot it. Better one bird in
    the hand than ten in the air. - the Wise from Antrim -

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    * Origin: fsxNet Usenet Gateway (21:1/5)